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This question already has an answer here:

I have already proven that $\left(1+\frac 1{n}\right)^n\leq \sum_{k=0}^{n}{\frac{1}{k!}}$ . Does that help me in any way? I am stumped...

Any suggestions/hints?

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marked as duplicate by Jyrki Lahtonen, Xander Henderson, Shailesh, YuiTo Cheng, Leucippus Jun 8 at 0:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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If $n,N\in\mathbb N$, with $n\geqslant N$, then\begin{align}\left(1+\frac1n\right)^n&=\sum_{k=0}^n\binom nk\frac1{n^k}\\&\geqslant\sum_{k=0}^N\binom nk\frac1{n^k}\\&=1+1+\frac{n-1}{2n}+\frac{(n-1)(n-2)}{3!n^2}+\cdots+\frac{(n-1)(n-2)\ldots\bigl(n-(N-1)\bigr)}{N!n^{N-1}}\\&=1+1+\frac1{2!}\left(1-\frac1n\right)+\frac1{3!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\cdots\\&\phantom{=\ }+\frac1{N!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots\left(1-\frac{N-1}n\right).\end{align}Therefore,\begin{align}\left(1+\frac1n\right)^n&\geqslant\lim_{n\to\infty}1+1+\cdots+\frac1{N!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\frac1{2!}\left(1-\frac1n\right)\cdots\left(1-\frac{N-1}n\right)\\&=1+1+\frac1{2!}+\cdots+\frac1{N!}\end{align}and so$$\lim_{n\to\infty}\left(1+\frac1n\right)^n\geqslant1+1+\frac1{2!}+\cdots+\frac1{N!}.$$Since this takes place for every $N\in\mathbb N$,$$\lim_{n\to\infty}\left(1+\frac1n\right)^n\geqslant\sum_{k=0}^\infty\frac1{k!}.$$

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  • $\begingroup$ $\begin{align}\left(1+\frac1n\right)^n&\geqslant\lim_{n\to\infty}1+1+\cdots+\frac1{N!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\frac1{2!}\left(1-\frac1n\right)\cdots\left(1-\frac{N-1}n\right)\\&=1+1+\frac1{2!}+\cdots+\frac1{N!}\end{align}$ How do you get that Inequality? $\endgroup$ – ParabolicAlcoholic Jun 8 at 11:04
  • $\begingroup$ No. I did no use that. $\endgroup$ – José Carlos Santos Jun 8 at 11:09
  • $\begingroup$ But how do you know that? $\endgroup$ – ParabolicAlcoholic Jun 8 at 11:27
  • $\begingroup$ How do I know what? Are you talking about the first inequality of my proof? $\endgroup$ – José Carlos Santos Jun 8 at 11:36
  • $\begingroup$ Ok, sorry, I misinterpreted somthing. It's clear now. $\endgroup$ – ParabolicAlcoholic Jun 8 at 11:45
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I don't know if this is what you're looking for but:

let exp(x) be the function that satisfies the D.E.: $$\frac{d}{dx}\exp(x) = \exp(x),\qquad \exp(0) = 1$$ hence if we do a Taylor expansion of this function we get: $$\exp(x) = \sum_{k=0}^\infty{\frac{x}{k!}}$$ using the limit definition of the derivative with a constant to the power x: $$\frac{d}{dx}\left(k^x\right) = \lim \limits_{h \to 0}\left(\frac{k^{x+h} - k^{x}}{h}\right) = \lim \limits_{h \to 0}\left(k^{x}\left(\frac{k^{h} - 1}{h}\right)\right) = T\cdot k^{x} $$ where $T = \lim \limits_{h \to 0}\left(\frac{k^{h} - 1}{h}\right)$ is constant with respect to x. Let e be the number such that $T = 1$, trivially, $e^x = \exp(x)$. rearranging our formula for T: $$\lim \limits_{h \to 0}\left[\frac{e^{h} - 1}{h}\right] = 1 \Rightarrow \lim \limits_{h \to 0}\left[e =\left(1+h\right)^\frac{1}{h}\right]$$ letting $u = \frac{1}{h}$: $$e =\lim \limits_{u \to \infty}\left[\left(1+\frac{1}{u}\right)^u\right]$$ hence: $$\lim \limits_{x \to \infty}\left[\left(1+\frac{1}{x}\right)^x\right] = e = \exp(1) = \sum_{k=0}^\infty{\frac{1}{k!}}$$

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For all $n\in\mathbb N$ $$\frac n{n+1}\le\ln\left(1+\frac1n\right)^n\le1.$$

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